[MiNT] Prusage(), Signals

[Ingo Schmidt <ich@der-ingo.de>]

Hi!

Two questions:

Prusage():

r[0] = process time in kernel
r[1] = process time in own program code

Does this mean I have to add those two times to get the overall time
the process has been running? I guess so.

And what does "process time in kernel" mean? Do VDI and AES also count
in here? E.g. if a VDI routine took a long time I will see that in
r[0]?


Signals:

Imagine the following situation (process A und process B)

A: Pexec(B)
A: Psigpause()
B: Pkill(A, SIGCONT)
B: Psigpause()
A: Pkill(B, SIGCONT)
A: Psigpause()
B: Pkill(A, SIGCONT)

and so on...

This looks very similar to Pmsg() in mode 2 (send message and then
wait for one). Andideally this will work, too. But since we have
multitasking, the following might occur:

A: Pexec(B)
   (now the OS switches tasks to process B, so A has not executed Psigpause() yet)
B: Pkill(A, SIGCONT)
B: Psigpause()
A: Psigpause()

So this is my question:
Will I have a deadlock now? Or does A immediately wake up again,
because the earlier SIGCONT signal was buffered? I read the docs, but
they are ambiguous here!! It is not celar to me what happens with the
signals.

I want to know this because I want to use the above scenario as a
(bad) replacement for Pmsg() under MagiC. But of course I do not want
deadlocks...


Cheers, Ingo =;->